Question

The values of $\alpha$ for which the point $(\alpha-1, \alpha+1)$ lies in the larger segment of the circle $x^{2}+y^{2}-x-y-6=0$ made by the chord whose equation is $x+y-2=0$ is

• A. $ -1<\alpha <1 $
• B. $ 1<\alpha <\infty $
• C. $ -\infty <\alpha <-1 $
• D. $ \alpha \le 0 $

Answer 1

Answer: (A)

The given circle
$S(x, y) \equiv x^{2}+y^{2}-x-y-6=0$ ... (i)
has centre at $C \equiv\left(\frac{1}{2}, \frac{1}{2}\right)$
According to the given conditions, the given point $P(\alpha-1, \alpha+1)$ must lie inside the given circle.
$i . e.S(\alpha-1, \alpha+1)< 0$
$\Rightarrow (\alpha-1)^{2}+(\alpha+1)^{2}-(\alpha-1)-(\alpha+1)-6<0$
$\Rightarrow \alpha^{2}-\alpha-2<0$ i.e.,$(\alpha-2)(\alpha+1)< 0$
$\Rightarrow -1< \alpha < 2$
[using sign - scheme from algebra] ... (ii)
and also $P$ and $C$ must lie on the same side of the line.

$L(x, y)=x+y-2=0$
i.e.,$L\left(\frac{1}{2}, \frac{1}{2}\right)$ and $L(\alpha-1, \alpha+1)$ must have the same sign.
Now, since $L\left(\frac{1}{2}, \frac{1}{2}\right)=\frac{1}{2}+\frac{1}{2}-2<0$
therefore, we have
$L(\alpha-1, \alpha+1)=(\alpha-1)+(\alpha+1)-2<0$
$\Rightarrow \alpha<1 $ ... (iv)
Ineqalities (ii) and (iv) together give the permissible values of $\alpha$ as $-1 < \alpha < 1$.